3.775 \(\int (a+b \cos (c+d x)) (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=61 \[ \frac {(a C+b B) \tan (c+d x)}{d}+\frac {(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*(B*a+2*C*b)*arctanh(sin(d*x+c))/d+(B*b+C*a)*tan(d*x+c)/d+1/2*a*B*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3029, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {(a C+b B) \tan (c+d x)}{d}+\frac {(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

((a*B + 2*b*C)*ArcTanh[Sin[c + d*x]])/(2*d) + ((b*B + a*C)*Tan[c + d*x])/d + (a*B*Sec[c + d*x]*Tan[c + d*x])/(
2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\int (a+b \cos (c+d x)) (B+C \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\int \left (a B+(b B+a C) \cos (c+d x)+b C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (2 (b B+a C)+(a B+2 b C) \cos (c+d x)) \sec ^2(c+d x) \, dx\\ &=\frac {a B \sec (c+d x) \tan (c+d x)}{2 d}+(b B+a C) \int \sec ^2(c+d x) \, dx+\frac {1}{2} (a B+2 b C) \int \sec (c+d x) \, dx\\ &=\frac {(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d}-\frac {(b B+a C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac {(a B+2 b C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(b B+a C) \tan (c+d x)}{d}+\frac {a B \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 75, normalized size = 1.23 \[ \frac {a B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a C \tan (c+d x)}{d}+\frac {b B \tan (c+d x)}{d}+\frac {b C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(a*B*ArcTanh[Sin[c + d*x]])/(2*d) + (b*C*ArcTanh[Sin[c + d*x]])/d + (b*B*Tan[c + d*x])/d + (a*C*Tan[c + d*x])/
d + (a*B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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fricas [A]  time = 0.43, size = 96, normalized size = 1.57 \[ \frac {{\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a + 2 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/4*((B*a + 2*C*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B*a + 2*C*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1)
+ 2*(B*a + 2*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [B]  time = 0.36, size = 151, normalized size = 2.48 \[ \frac {{\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/2*((B*a + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (B*a + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
B*a*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1
/2*c) + 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 0.30, size = 86, normalized size = 1.41 \[ \frac {a C \tan \left (d x +c \right )}{d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B b \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

1/d*a*C*tan(d*x+c)+1/2/d*a*B*sec(d*x+c)*tan(d*x+c)+1/2/d*a*B*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*b*ln(sec(d*x+c)+t
an(d*x+c))+1/d*B*b*tan(d*x+c)

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maxima [A]  time = 0.34, size = 95, normalized size = 1.56 \[ -\frac {B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, C b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C a \tan \left (d x + c\right ) - 4 \, B b \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/4*(B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 2*C*b*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*C*a*tan(d*x + c) - 4*B*b*tan(d*x + c))/d

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mupad [B]  time = 2.77, size = 104, normalized size = 1.70 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,a+2\,B\,b+2\,C\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,b-B\,a+2\,C\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B\,a+2\,C\,b\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x)^4,x)

[Out]

(tan(c/2 + (d*x)/2)*(B*a + 2*B*b + 2*C*a) - tan(c/2 + (d*x)/2)^3*(2*B*b - B*a + 2*C*a))/(d*(tan(c/2 + (d*x)/2)
^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(B*a + 2*C*b))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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